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The formula we use to calculate gravitational time dilation is $$T_{dilated}=\sqrt{1-\frac{2Gm}{Rc^2}} \cdot T_{without gravity}.$$

But this formula has a problem that is, the gravitational time dilation for distances smaller than $\frac{2Gm}{c^2}$ we have an imaginary time dilation. So how can we calculate time dilation near the surface of an object?

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    $\begingroup$ What do you mean "near the surface" - what "objects" are you dealing with where that distance is not firmly inside the object? $\endgroup$
    – ACuriousMind
    Oct 26, 2019 at 12:57

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The $R$ in that equation is the distance to the center of the massive object. For the Earth:

$$ \frac{2GM_\mathrm{Earth}}{c^2} \approx 9\,\mathrm{mm}, $$

and for the Sun:

$$\frac{2GM_\mathrm{Sun}}{c^2} \approx 3\,\mathrm{km}. $$

As @ACuriousMind implies in a comment, that equation only works outside of a massive object. For any real object you would be well inside before hitting the $R = 2Gm/c^2$ mark.

It might help to see what's going on, if you know where that equation came from. That expression for gravitational time dilation comes from the Schwarzschild solution to Einstein's field equations from General Relativity.

The Schwarzschild metric tell us about the space-time in a vacuum near a stationary, spherically symmetric mass. That it's a vacuum solution is why whether you are inside or outside the massive object matters. If you are 1 km from the center of the Sun you are not in a vacuum, so you need a different mathematical description of what's going on.

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According to this information, black holes can exist but never form. Say, a neutron star containing about three solar masses and having a diameter of 6000 km begins to collapse. Since it cannot collapse faster than the speed of light, this will take about 10 milliseconds. For the first 9 milliseconds, everything goes according to schedule, but then the gravitational time dilution sets in. Seen from the star everything proceeds as normal, but seen from the outside the collapse is going slower and slower and when the star has a diameter of about 15 km, it will only collapse at a snails pace. It will never reach singularity and will still emit electromagnetic radiation. However, due to the time dilation, this radiation will have an extremely long vawelength seen from the outside. Even gammarays will be twisted so much that they are not easily detected. So the star may appear as a black hole, but in fact is not.

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    $\begingroup$ The question isn't specifically about black holes. $\endgroup$
    – D. Halsey
    Mar 7, 2023 at 13:53

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